Topscore Test 2 GC Question #68

Someone posted this question before.. It is a weird question..
Lets hope someone can explain it this time..
http://forums.studentdoctor.net/showthread.php?t=824970

bittersweet008x said:

What volume of HCl was added if 20 mL of 1M NaOH is titrated with 1 M HCl to produce a pH=2?

The answer is 20.4

& the solution on topscore is just confusing me more!

help please?

Click to expand...

pH = 2
[H+] = 10^(-2)
[H+] = 0.01 M

Since HCl + NaOH -> H2O + NaCl, it's a neutralization reaction.

20 mL 1M HCl + 20 mL 1M NaOH = 40 mL of water

If we keep adding HCl, the pH goes down as HCl will dissociate into H+ and Cl-.

How much extra HCl do we add?
Given:
Volume ~ 40 mL
Concentration = 0.01 M
moles = ?

c = moles/volume
moles = (0.040L)(0.01M)
moles = 0.0004 mol

We know the concentration of HCl is 1M.
So the amount of 1M HCl needed to add to the solution is..

c=moles/volume
volume = moles/concentration
volume = 0.0004 mol/1.0M
volume = 0.0004 L = 0.4 mL


Total Volume of HCl added = neutralizing volume + volume needed to lower pH
= 20 mL + 0.4 mL
= 20.4 mL

Hope this helps...

bittersweet008x said:

What volume of HCl was added if 20 mL of 1M NaOH is titrated with 1 M HCl to produce a pH=2?

The answer is 20.4

& the solution on topscore is just confusing me more!

help please?

Click to expand...

bharat008 said:

Someone posted this question before.. It is a weird question..
Lets hope someone can explain it this time..
http://forums.studentdoctor.net/showthread.php?t=824970

Click to expand...

I got this...

20 mL of 1M NaOH is titrated with 1 M HCl to produce a pH=2?

To start off we have to see what the titration curve would look like. Since they are both strong monoprotic acid and base we know the equivalence point is around 7. We are making an acidic solution so the pH is less than the equivalence point meaning that we are going to add more acid.
We first add 20 ml of HCl to neutralize the acid that we have and get to pH 7.

Now our final volume of our hypothetical pH 2 solution is 40 + x (since you added 20 ml HCL to 20 ml NaOH). X is however more 1 M HCl you have to add to get an acidic solution. Now we are going to deduce the concentration of the final pH 2 solution which would simply be [10^-2] = .1

We want to see how much should be added of that 1 M HCl.

Our equation now becomes:

(1)(x) = (.01)(40 + x)
x = .4 + .01x
.99x = .4
x = .404

So the total volume of HCl that was added was:
20 (first batch we added to neutralize the acid) + .404 (second batch to get to pH 2) = 20.4

Hope that helps

Damn never mind then

Both are awesome explanations. I got this wrong on topscore as well...